Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $x = \dfrac{-8k + 64}{-7k - 7} \times \dfrac{k + 1}{k^2 + k - 72} $
Explanation: First factor the quadratic. $x = \dfrac{-8k + 64}{-7k - 7} \times \dfrac{k + 1}{(k - 8)(k + 9)} $ Then factor out any other terms. $x = \dfrac{-8(k - 8)}{-7(k + 1)} \times \dfrac{k + 1}{(k - 8)(k + 9)} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac{ -8(k - 8) \times (k + 1) } { -7(k + 1) \times (k - 8)(k + 9) } $ $x = \dfrac{ -8(k - 8)(k + 1)}{ -7(k + 1)(k - 8)(k + 9)} $ Notice that $(k + 1)$ and $(k - 8)$ appear in both the numerator and denominator so we can cancel them. $x = \dfrac{ -8\cancel{(k - 8)}(k + 1)}{ -7(k + 1)\cancel{(k - 8)}(k + 9)} $ We are dividing by $k - 8$ , so $k - 8 \neq 0$ Therefore, $k \neq 8$ $x = \dfrac{ -8\cancel{(k - 8)}\cancel{(k + 1)}}{ -7\cancel{(k + 1)}\cancel{(k - 8)}(k + 9)} $ We are dividing by $k + 1$ , so $k + 1 \neq 0$ Therefore, $k \neq -1$ $x = \dfrac{-8}{-7(k + 9)} $ $x = \dfrac{8}{7(k + 9)} ; \space k \neq 8 ; \space k \neq -1 $